Answer
$${y^2} = - \frac{2}{3}{e^x} + C{e^{4x}}$$
Work Step by Step
$$\eqalign{
& yy' - 2{y^2} = {e^x} \cr
& y' - \frac{{2{y^2}}}{y} = {e^x}\left( {\frac{1}{y}} \right) \cr
& y' - 2y = {e^x}{y^{ - 1}} \cr
& {\text{The differential equation is written in the form}} \cr
& y' + P\left( x \right)y = Q\left( x \right){y^n} \cr
& y' - 2y = {e^x}{y^{ - 1}} \to P\left( x \right) = - 2,{\text{ }}Q\left( x \right) = {e^x},{\text{ }}n = - 1 \cr
& {\text{From the book we have that the general solution of a Bernoulli }} \cr
& {\text{equation is:}} \cr
& {y^{1 - n}}{e^{\int {\left( {1 - n} \right)P\left( x \right)dx} }} = \int {\left( {1 - n} \right)Q\left( x \right){e^{\int {\left( {1 - n} \right)P\left( x \right)dx} }}dx} + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Find }}{e^{\int {\left( {1 - n} \right)P\left( x \right)dx} }} \cr
& {e^{\int {\left( {1 - n} \right)P\left( x \right)dx} }} = {e^{\int {\left( {1 + 1} \right)\left( { - 2} \right)dx} }} = {e^{ - 4\int {dx} }} = {e^{ - 4x}} \cr
& {\text{Substituting }}{e^{\int {\left( {1 - n} \right)P\left( x \right)dx} }},{\text{ }}n,{\text{ and }}Q\left( x \right){\text{ into }}\left( {\bf{1}} \right) \cr
& {y^{1 + 1}}{e^{ - 4x}} = \int {\left( {1 + 1} \right)\left( {{e^x}} \right){e^{ - 4x}}dx} + C \cr
& {y^2}{e^{ - 4x}} = \int {2{e^{ - 3x}}dx} + C \cr
& {\text{Integrating}} \cr
& {y^2}{e^{ - 4x}} = - \frac{2}{3}{e^{ - 3x}} + C \cr
& {\text{Solve for }}y \cr
& {y^2} = - \frac{2}{3}\frac{{{e^{ - 3x}}}}{{{e^{ - 4x}}}} + \frac{C}{{{e^{ - 4x}}}} \cr
& {y^2} = - \frac{2}{3}{e^x} + C{e^{4x}} \cr} $$
